I'm not so sure this question is "well defined". What I mean is that a priori F_2 could sit inside G=SL_{2}(Z) in two different ways so the index of the subgroups are not the same. An evidence of what I'm saying are the answers below. Even though all of them construct normal subgroups of index 12, one of the quotients is cyclic while the other is not. I believe the actual question you had in mind is: How to show that the commutator has index 12, and moreover why is it isomorphic to F_2? If you can prove that every embedding of F_2 in G has the same index then original question makes sense.
–
Guillermo MantillaOct 27 '10 at 0:02

1

I think the subgroup $\Gamma_1(4)$ is another example, and this one isn't even normal. It's torsion-free so acts freely on the usual tree, so is free, and then the Euler characteristic argument says it's rank 2.
–
Kevin BuzzardOct 27 '10 at 0:14

12

@Guillermo Mantilla: there will be plenty of free groups of rank 2 of infinite index in SL(2,Z) (e.g. a free group of rank 2 contains a free group of rank 3 with finite index, and this contains a free group of rank 2 with infinite index). All the finite index ones will have index 12 though, by the Euler characteristic argument.
–
Kevin BuzzardOct 27 '10 at 0:16

The answers bring out lots of nice viewpoints, but I think it would also be helpful to give more explicit references, going back not just to Serre (whose viewpoint I like very much) but farther to the origins which probably occur in work of Schreier and the Neumanns. Certainly the structure of the modular group has been understood for a long time, in the general context of free products with amalgamation and their subgroups.
–
Jim HumphreysOct 27 '10 at 18:55

9 Answers
9

To elaborate on Qiaochu's answer. The subgroup generated by the two matrices

$$\left[ \begin{array}{cc} 1 & 2 \\\ 0 & 1 \end{array} \right]$$
and

$$\left[ \begin{array}{cc} 1 & 0 \\\ 2 & 1 \end{array} \right]$$
is the Sanov subgroup. It consists, by an exercise in Kargapolov-Merzlyakov, of matrices of the form
$$\left[ \begin{array}{cc} 4k+1 & 2l \\\ 2m & 4n+1 \end{array} \right]$$ and det=1. The congruence subgroup $\Gamma(2)$ consists of matrices of the form $$\left[ \begin{array}{cc} 2k+1 & 2l \\\ 2m & 2n+1 \end{array} \right]$$ and det=1. Those matrices from $\Gamma(2)$ and not in the Sanov subgroup have the form $$\left[ \begin{array}{cc} 4k+3 & 2l \\\ 2m & 4n+3 \end{array} \right].$$ Taking the product of any two such matrices gives us a matrix from the Sanov subgroup. So the Sanov subgroup has index 2 in $\Gamma(2)$, and index 12 in $SL_2(\mathbb Z)$.

There have already been a lot of answers to this question and I'm not going to go over the details of a proof that it is true. But I want to say something about the real strength of any of these standard arguments. The arguments generally have one of three starting points: (1) That $\text{PSL}(2,\mathbb{Z})$ is isomorphic to $C_2 * C_3$, a free product of cyclic groups; (2) The group action of $\text{SL}(2,\mathbb{Z})$ on the hyperbolic plane or upper half plane; or (3) A group action of $\text{SL}(2,\mathbb{Z})$ on trees. The first starting point arguably dodges the question, because obtaining that the whole group is a free product is closely related to finding a free subgroup, The other two starting points are "geometric". But geometry is ultimately a human description of some other, more formal mode of reasoning. So what are the geometric arguments really saying?

It is important to realize that the statement that a group $G$ is free is ultimately a uniqueness statement for words in $G$ in its claimed free generators. More precisely, if $G$ is freely generated by $a$ and $b$, that says that every element of $G$ has unique reduced word in $a$ and $b$ and their inverses. In the case of this subgroup of $\text{SL}(2,\mathbb{Z})$ or $\text{PSL}(2,\mathbb{Z})$, the uniqueness property ultimately comes from and generalizes another, more familiar uniqueness property: The uniqueness of the Euclidean algorithm with positive remainders.

Suppose that you have two coprime positive integers $a$ and $b$ and you are allowed to subtract $a$ from $b$, or $b$ from $a$, but you are not allowed to create negative integers. Then there is a unique path from $(a,b)$ to $(1,1)$. The same statement in reverse, in linear algebra form, says that there is a unique word in the matrices
$$\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \qquad \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$$
which, when applied to the vector $(1,1)$, gives you the vector $(a,b)$. Thus, these two matrices generate a free semigroup.

The uniqueness property is established by induction, and the obvious fact that if $a < b$ then you can only subtract $a$ from $b$, and vice versa. Obvious though it may be, this type of fact is what lets you build the tree on which $\text{SL}(2,\mathbb{Z})$ acts. Or, in the hyperbolic geometry version of the argument, the inequality $a < b$ is associated with the sides of the ideal polygons that eventually yield the same tree. In the upper half plane model, the rational numbers $a/b$ with $a < b$ are a ray in the real numbers, and their convex hull in the hyperbolic plane is a half hyperbolic plane whose boundary is the geodesic connecting $\infty$ to 1. Uniqueness says that the orbit of this geodesic under the above semigroup consists of disjoint geodesics.

Well $SL_2(\mathbb{Z})\cong \mathbb{Z}/6*_{\mathbb{Z}/2} \mathbb{Z}/4$. $SL_2(\mathbb{Z})$ acts on a tree $T$. Then you can consider its abelianization. It is $\mathbb{Z}/12$. So the commutator subgroup $\Gamma$ has index 12. It is free, as it acts freely on the same tree $T$. See for example Serre, trees for details.

EDIT: One can also investigae the rank of the free group. Thatfor one has to consider the quotient graphs $SL_2(\mathbb{Z})\backslash T$ and $\Gamma\backslash T$. The first Graph is a interval, where the stabilizers of representatives of one endpoint are isomorphic to $\mathbb{Z}/6$ (resp. $\mathbb{Z}/4$ for the other endpoint). The stabilizer of a representative of the inverval is $\mathbb{Z}/2$.

Now $\Gamma$ acts freely on $T$ and $[SL_2(\mathbb{Z}):\Gamma]=12$. So the first endpoint (which is a $SL_2(\mathbb{Z})$-orbit) consists of $12/6=2$ $\Gamma$-orbits and the second endpoint consists of $12/4=3$ $\Gamma$-orbits. Analogously the edge consists of $6$ $\Gamma$-orbits.

So the Euler Characteristic of $\Gamma\backslash T$ is $2+3-6=-1$, so this graph is homotopy equivalent to a wedge of $2$ circles and hence its fundamental group (which is isomorphic to $\Gamma$ by Serre, trees or covering theory) is $F_2$.

Kevin: by construction of the Bass--Serre tree, any point stabiliser is conjugate into Z/4 or Z/6. Now note that both of these subgroups inject into the abelianisation, and you're done. This is a special feature of this situation.
–
HJRWOct 27 '10 at 14:20

This is an elementary version of my Euler number argument. I like it.
–
Johannes EbertOct 28 '10 at 18:24

1

Henrik, you certainly don't mean that 'this graph is homotopy equivalent to a wedge of 2 spheres'. Circles, perhaps? In fact, as in my answer below, it's the theta graph.
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HJRWNov 2 '10 at 15:07

Start with the formula for $SL_2(\mathbb Z)$ in HenrikRueping's answer. Divide by the central $\mathbb Z/2$ to conclude that $PSL_2(\mathbb Z)$ is $\mathbb Z/3*\mathbb Z/2$. The kernel of abelianization for $SL_2(\mathbb Z)$ is the same as the kernel of abelianization for $PSL_2(\mathbb Z)$. The kernel of $G*H \to G\times H$ is always a free group with basis $ghg^{-1}h^{-1}$, $g\ne 1$, $h\ne 1$.

The main problem is the rank of the kernel, it was addressed above. The fact that the derived subgroup does not contain the center, and hence the same for SL and PSL, follows from the fact that it is free (in a nilpotent group, the derived subgroup quite often contains the center). So this is not a better solution than Henrik's.
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Mark SapirOct 27 '10 at 0:40

Sure. I just thought that this alternative reason might be considered more elementary; and it does give an explicit basis for the free group.
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Tom GoodwillieOct 27 '10 at 1:13

@Tom: The good point of your argument is that it avoids the Bass-Serre theory which is indeed not necessary. But in order to show that the kernel misses the center, and hence it is free, you are using the fact that it is free (or at least torsion-free). That makes your argument somewhat circular. Also although your argument indeed shows that the kernel is generated by two elements as a normal subgroup it says nothing about the number of subgroup generators. It is quite possible that your argument can be fixed.
–
Mark SapirOct 27 '10 at 3:08

2

@Mark: It's not circular. Let $A,B,C$ be abelian groups with injective homomorphisms $C\to A$ and $C\to B$. Let $G$ be $A*_CB$. Let $K$ be the kernel of the abelianization map $G\to A\oplus_C B$. The image of the evident homomorphism $C\to G$ is central. Let $PG$ be the quotient, and let $PA$ and $PB$ be the cokernels of $C\to A$ and $C\to B$. Thus $PG$ is $PA*PB$. Let $PK$ be the kernel of the abelianization map $PG\to PA\oplus PB$. The homomorphism $G\to PG$ induces another $K\to PK$. The latter is both surjective and injective by diagram-chasing.
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Tom GoodwillieOct 27 '10 at 4:38

@Mark: And the kernel of the map from free product to cartesian product is always free, with rank equal to (order-1)x(order-1).
–
Tom GoodwillieOct 27 '10 at 4:39

Maybe this works and maybe it doesn't: the index of the congruence subgroup $\Gamma(2)$ in $\text{SL}_2(\mathbb{Z})$ is $|\text{SL}_2(\mathbb{F}_2)| = 6$, and $\Gamma(2)$ contains a free subgroup of index $2$ generated by $\left[ \begin{array}{cc} 1 & 2 \\\ 0 & 1 \end{array} \right]$ and $\left[ \begin{array}{cc} 1 & 0 \\\ 2 & 1 \end{array} \right]$, which follows from the fact that its action on the hyperbolic plane is free.

Thanks for the question!; I asked this myself several times, without definite answer. But now I got it. We know from Henrik Rüpings answer that the kernel $K$ of $Sl_2 (Z) \to Z/12$ is a free group. Thus it remains to determine the rank. There is a notion of Euler characteristic for groups (it has rational values), see Brown's book, cohomology of groups. Theorem IX 6.3 (or better IX 7.3 b) of that book states that $\chi(Sl_2 (Z))= \chi (Z/12) \chi(K)$ in our case. The Euler number of $Z/12$ is $\frac{1}{12}$, the Euler number of $Sl_2 (Z)$ is $-\frac{1}{12}$, so the Euler number of $K$ is $-1$. But the Euler number of $F_n$ is $1-n$, so $n=2$, and you are done.

I would like to see a more elementary answer, though; one should be able to see this using the action of the upper half plane; then glue $12$ fundamental domains together in the correct way and see that the result is $C \setminus \{0,1\}$. But I never managed to do this properly.

@José Figueroa-O'Farrill: of course it is: a theorem of Harder relates the Euler numbers of arithmetic groups to zeta-values.
–
Johannes EbertOct 27 '10 at 0:04

1

Yes, I was too quick. The Sanov subgroup is the kernel of a homomorphism onto a different group of order 12.
–
Mark SapirOct 27 '10 at 0:05

1

@Jose: most of this story is in Brown's book (I do not know Serre's "Trees" that well). In one section, he explains how groups acting on trees are decomposed into amalgamated products and how the Mayer-Vietoris sequence can be used in group homology calculations, which is of course a rather elusive situation. He discusses the example of $Sl_2(Z)$, following Serre.
–
Johannes EbertOct 27 '10 at 21:24

I'm late to the party, but here's how I think about this fact. The approach I will present is perhaps slightly higher tech than some, but has the advantage that it's a completely general way of computing a free finite-index subgroup of a virtually free group.

To keep things simple, I'll do $PSL_2(\mathbb{Z})$, and then explain how to do $SL_2(\mathbb{Z})$ at the end.

As you presumably know, $\Gamma=PSL_2(\mathbb{Z})\cong \mathbb{Z}/2*\mathbb{Z}/3$. You should think of it as the fundamental group of a graph of groups $G$, with one edge and two vertices, one of which, $u$, is labelled by $\mathbb{Z}/2$ and one, $v$, labelled by $\mathbb{Z}/3$.

Just as in the case of ordinary topology, any (conjugacy class of a) subgroup $H\subseteq\Gamma$ corresponds to a covering map $\widehat{G}\to G$. (This is a trivial consequence of Bass--Serre theory: take $\widehat{G}=T/H$ where $T$ is the Bass--Serre tree.) Coverings of graphs of groups are just like coverings of graphs, except that sometimes a neighbourhood looks like a quotient by a non-free action of a group, in which case you have to remember point stabilisers (cf. orbifolds). The subgroup $H$ is free if and only if every vertex of $\widehat{G}$ is labelled by the trivial group, in which case we can think of $\widehat{G}$ as a genuine graph and $H$ as its fundamental group.

To prove that $PSL_2(\mathbb{Z})$ has a free subgroup of index $6$, we now just have to construct a covering map $\widehat{G}\to G$ of degree $6$, where $\widehat{G}$ is a genuine graph. But this is easy. To do this, take three vertices $u_1,u_2,u_3$ of valence two and two vertices $v_1,v_2$ of valence three. Now glue them together however you like, such that each edge adjoins one $u_i$ and one $v_j$. (I think there are exactly two ways of doing this, up to permuting the $i$'s and $j$'s.) The covering map $\widehat{G}\to G$ sends $u_i\mapsto u$ and $v_j\mapsto v$, and each edge goes to the unique edge of $G$. Whichever way you do the gluing, the graph $\widehat{G}$ has rank two, so the subgroup is isomorphic to $F_2$.

It's also easy to see that these subgroups are the minimal-index torsion-free subgroups: any covering of $G$ by a genuine graph $\widehat{G}$ must have even degree, so that the vertices above $u$ have trivial label; and it must have degree divisible by three, so that the vertices above $v$ have trivial label.

Exactly the same argument can be made to work for $SL_2(\mathbb{Z})$, which you of course should think of as $\mathbb{Z}/4*_{\mathbb{Z}/2}\mathbb{Z}/6$. Indeed, if I have it right, these techniques should prove the following theorem, which is presumably standard, although I don't know a reference. (Warning: I haven't thought about this very hard.)

Theorem: If $\Gamma=A*_CB$ for $A,B$ finite, then a minimal-index free subgroup is of index $d=\mathrm{lcm}(|A|,|B|)$, and of rank

$1-d\left(\frac{1}{|A|}+\frac{1}{|B|}-\frac{1}{|C|}\right)$.

Note that this technique is powerful enough to let you compute all the conjugacy classes of all the subgroups of a given index, if you want. (Update: though there are some subtleties if you really want to do this; see below.)

(At the moment what I'm seeing doesn't quite square with some of the other answers/comments: I get one normal free subgroup of $PSL_2(\mathbb{Z})$ of index six, and one conjugacy class of non-normal free subgroups of index six, corresponding to a regular covering of $G$ and an irregular covering, respectively; this seems at odds with Mark's remark about different maps to $\mathbb{Z}/12$. But it's late, so I'm not going to figure out what's going on now.)

Update

I had forgotten an important subtlety. A covering map of graphs of groups $\widehat{G}\to G$ includes an extra piece of information, viz:

For each vertex $\hat{v}$ of $\widehat{G}$ with image $v$ in $G$, and for each edge $e$ incident at $v$, a covering map specifies a bijection between the set of edges of $\widehat{G}$ incident at $\hat{v}$ that map to $e$ and the set of double cosets

$\widehat{G}_{\hat{v}}\backslash G_v/G_e$

(where $G_v$ is the group labelling $v$, etc). As such bijections always exist, this doesn't arise if you want to prove the existence of a subgroup. But the correct notion of equivalence for covering maps takes these bijections into account, so two different covering maps $\widehat{G}\to G$ may have the same underlying graph-map. This happens for the commutator subgroup and the Sanov subgroup: in both cases, the underlying graph is the first subdivision of the theta graph.

I think the Sanov subgroup described by Mark, and the commutator subgroup, are both good candidates for (distinct) index 6 normal subgroups of PSL(2,Z). The [image of the] Sanov subgroup [in PSL(2,Z)] is just Gamma(2) in the PSL world, so the quotient is PSL(2,Z/2Z)=S_3, and the commutator subgroup has quotient Z/6Z. Does this help unconfuse matters?
–
Kevin BuzzardOct 27 '10 at 13:28

Kevin - well, your comment persuades me that it's my mistake and not Mark's, so it does at least narrow down the source of the confusion!
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HJRWOct 27 '10 at 14:12

May be the discussion is out of date, but I like to present a simple way to show that the subgroup S generated by
$$
\left(\begin{array}{cc} 1 & 2\cr
0 & 1 \end{array}\right)
$$ and
$$
\left(\begin{array}{cc} 1 & 0\cr
2 & 1 \end{array}\right)
$$ has index 12 in $SL(2,Z)$.
First of all Mark Sapir noted that by Kargopolov S is a group of matrices
$$
\left(\begin{array}{cc} 1+4k_1 & 2n_1\cr
2n_2 & 1+4k_2 \end{array}\right)
$$

Lemma. Let $G$ be a group,
$H$ be a subgroup of $G$, $N$ be a normal subgroup of $G$, that is subgroup of $H$.
Then ind(G:H)=ind(G/N:H/N).

A remark on the original question: it follows from either the results of Gilman/Miasnikov/Osin or of Richard Aoun (for linear groups) that the subgroup generated by a random (in an appropriate sense) pair of elements in $SL(2, \mathbb{Z})$ is free. It can also be shown that such a subgroup will (with very high probability) be of infinite index in the modular group.